Problem: Is ${50334}$ divisible by $3$ ?
A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {50334}= &&{5}\cdot10000+ \\&&{0}\cdot1000+ \\&&{3}\cdot100+ \\&&{3}\cdot10+ \\&&{4}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {50334}= &&{5}(9999+1)+ \\&&{0}(999+1)+ \\&&{3}(99+1)+ \\&&{3}(9+1)+ \\&&{4} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {50334}= &&\gray{5\cdot9999}+ \\&&\gray{0\cdot999}+ \\&&\gray{3\cdot99}+ \\&&\gray{3\cdot9}+ \\&& {5}+{0}+{3}+{3}+{4} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first four terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${50334}$ is divisible by $3$ if ${ 5}+{0}+{3}+{3}+{4}$ is divisible by $3$ Add the digits of ${50334}$ $ {5}+{0}+{3}+{3}+{4} = {15} $ If ${15}$ is divisible by $3$ , then ${50334}$ must also be divisible by $3$ ${15}$ is divisible by $3$, therefore ${50334}$ must also be divisible by $3$.